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I have seen the notation $\int_M fe^{\mu}$ in some geometry books and I cannot even guess what $e^{\mu}$ might mean for a measure/form $\mu$ on the (symplectic) manifold $M$.

Any clarifications are appreciated.

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  • $\begingroup$ Where did you see this notation? $\endgroup$
    – user99914
    Commented Nov 10, 2014 at 5:42
  • $\begingroup$ @John I included an example from repn theory/symp geom. $\endgroup$
    – Philip M
    Commented Nov 10, 2014 at 6:27
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    $\begingroup$ See en.wikipedia.org/wiki/Emu $\endgroup$ Commented Nov 10, 2014 at 6:44
  • $\begingroup$ @JairTaylor what about Esigma? $\endgroup$
    – Philip M
    Commented Nov 10, 2014 at 6:50
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    $\begingroup$ My only thought is $e^\sigma = e^{(\sigma/\omega)} \omega$ for $\omega$ the canonical volume form, but I have never seen the notation before, nor studied symplectic geometry. Do none of the books define it? $\endgroup$ Commented Nov 10, 2014 at 7:12

2 Answers 2

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I found this remark in a book by Guillemin, Ginzburg, and Karshon:

It is convenient to work with the differential form (of mixed degree) $$\exp \omega=1+\omega+\frac{1}{2!}\omega\wedge\omega\dots .$$ With the convention that $\int_M\beta=0$ if the degree of $\beta$ is different than the dimension of $M$, Liouville measure is given by integration of $\exp \omega$.

This shows that my guess in the above comments was correct! However, I would appreciate if an expert would clarify why it is "convenient" to work with this differential form of mixed degrees.

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  • $\begingroup$ I guess because it's easier to write $e^\omega$ than $\frac{1}{7!} \omega\wedge\omega \wedge \dotsc \wedge\omega$. But that's just a guess. $\endgroup$ Commented Nov 10, 2014 at 22:21
  • $\begingroup$ @DanielFischer I don't think writing $e^{\omega}$ is so much easier than $\omega^n/n!$ that justifies its use. $\endgroup$
    – Philip M
    Commented Nov 10, 2014 at 22:59
  • $\begingroup$ This remind me of Chern character en.wikipedia.org/wiki/Chern_class $\endgroup$
    – user99914
    Commented Nov 11, 2014 at 2:58
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If $e^{\mu}=d\lambda$ for some measure $\lambda$ in $M$ then, it seems that $\mu=\ln(d\lambda)$. But here is not clear whether $\ln(d\lambda)$ is a measure.

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